- #91

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Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).##2x^6+3y^6=z^3##

It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).

We do not "see" it. What we see is that you invented out of the blue a certain number ##t##. What you should have done is telling us, that you set ##t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}## such that ##y=\frac{2}{3}t^3##. But now ##x^3\neq \frac{1}{2}-t^3##.We see that ##\frac{3y^3}{2} = \frac{1}{2} -x^3##

We see that ##x^3= \frac{1}{2} - t^3## and ##y= \frac{2t^3}{3}##

To correct this, we have to set ##t:=\sqrt[3]{\frac{3}{2}}y##. Now ##x^3= \frac{1}{2}-t^3##.

How that?If t is rational, then y is irrational.

If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.